Question

The real force ${}^{\prime }{F}^{\prime }$ acting on a particle of mass ${}^{\prime }{m}^{\prime }$ performing circular motion acts along the radius of circle ${}^{\prime }{r}^{\prime }$ and is directed towards the centre of circle. The square root of magnitudeof such force is ($T=$ periodic time)

• A. $\frac{2\pi }{T}\sqrt{mr}$
• B. $\frac{Tmr}{4\pi }$
• C. $\frac{2\pi T}{\sqrt{mr}}$
• D. $\frac{T^{2}mr}{4\pi }$

Answer 1

Answer: (A)

Force $F$ acting on a body of mass $m$ performing circular motion of radius $r$,
$F=\frac{m{v}^2}{r}$ (centripetal force) ...(i)
where, $v=$ velocity of the particle The time period of one complete cycle,
$T=\frac{\text{ perimeter of a circular path }}{\text{ velocity of body }}=\frac{2\pi r}{v}$
$\Rightarrow v=\frac{2\pi r}{T}$...(ii)
From Eqs. (i) and (ii), we get
$F=\frac{m}{r}{\left(\frac{2\pi r}{T}\right)}^2=mr{\left(\frac{2\pi }{T}\right)}^2$
or $\sqrt{F}=\frac{2\pi }{T}\sqrt{mr}$
So, option (a) is correct.